Pls help me with this question:As a result of this process offspring 1 and offspring 2 will have?

Free brainlest just thank my other questions

alexandr402 [8]

Answer:

ok and thank you for free point

6 0

2 years ago

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13. Would you expect oxygen to form a cation or anion? How many electrons would it gain or lose? Why?

netineya [11]

Answer: C)Anion, it would gain 2 electrons to satisfy the octet rule.

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electrons are filled according to Afbau's rule in order of increasing energies and thus the electronic configuration of oxygen with 8 electrons is

$O:8:1s^22s^22p^4$

The cation is formed by loss of electrons and anions are formed by gain of electrons.

In order to complete its octet and get stable, it gains 2 electrons and thus would form an anion.

$O^{2-}:10:1s^22s^22p^6$

7 0

3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant o

anzhelika [568]

Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

HNO2 <--> H+ NO2-
[]i 0.139M 0M 0M
Δ[] -x +x +x
[]f 0.139-x x x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.007M/0.139M = .0503 or

≈5.03% dissociation.

4 0

3 years ago

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fish give off the compound ammonia which has a pH above 7 to which class of compounds does ammonia belong

jasenka [17]

Answer:
The compound ammonia given by fish is alkaline

Explanation:
We can classify elements/compounds based on their pH values into three types:
acids: these are compounds having pH value lower than 7
neutral: these are compounds having pH value equal to 7
alkalies: these are compounds having pH values higher than 7

This is shown in the attached image

We are given that the pH of the compound ammonia generated by the fish is above 7.
According to the above explanation, compound ammonia would be an alkaline compound.

Hope this helps :)

5 0

3 years ago

How many grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate?

Korvikt [17]

Answer: There is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.

Explanation:

Chemical equation depicting reaction between sodium phosphate and aluminum carbonate is as follows.

$Al_{2}(CO_{3})_{3} + 2Na_{3}PO_{4} \rightarrow 2AlPO_{4} + 3Na_{2}CO_{3}$

As this equation contains same number of atoms on both reactant and product side. So, this equation is a balanced equation.

According to the equation, 2 moles of sodium phosphate is giving 3 moles of sodium carbonate.

Therefore, sodium carbonate formed by 5.3 moles of sodium phosphate is as follows.

$\frac{3}{2} \times 5.3 mol\\= 7.95 mol$

As number of moles is the mass of substance divided by its molar mass. So, mass of sodium carbonate ( molar mass = 105.98 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\7.95 mol = \frac{mass}{105.98 g/mol}\\mass = 842.54 g$

Thus, we can conclude that there is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.

8 0

2 years ago